| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Given conditional, find joint or marginal |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question requiring only direct application of P(A∩B) = P(B|A)×P(A), basic Venn diagram completion using given probabilities, and checking independence via P(A∩B) = P(A)×P(B). All parts follow standard textbook procedures with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A \cap B) = P(B | A) \times P(A) = \frac{3}{7} \times \frac{7}{10} = \frac{3}{10}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Venn diagram with four regions: | B1 | |
| \(P(A \cap B) = 0.3\), \(P(A \cap B') = 0.4\), \(P(A' \cap B) = \) value, outside \(= 0.1\) | B1 | |
| Need \(P(B) = 1 - 0.1 - 0.4 - P(A' \cap B)\); \(P(A' \cap B)\) found from \(P(B) - 0.3\) | ||
| \(P(A' \cap B') = 0.1\), \(P(A \cap B') = 0.4\), \(P(A \cap B) = 0.3\), \(P(A' \cap B) = 0.2\) | A1 ft | All four regions correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B) = 0.3 + 0.2 = 0.5\) | M1 | |
| \(P(B | A) = \frac{3}{7} \neq P(B) = 0.5\), so not independent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B | A) = \frac{3}{7} \approx 0.43 < P(B) = 0.5\) | M1 |
| Parents attending reduces probability of scoring, so Isobel should ask them not to attend | A1 |
# Question 2:
## Part (i)
| $P(A \cap B) = P(B|A) \times P(A) = \frac{3}{7} \times \frac{7}{10} = \frac{3}{10}$ | M1 A1 | |
## Part (ii)
| Venn diagram with four regions: | B1 | |
| $P(A \cap B) = 0.3$, $P(A \cap B') = 0.4$, $P(A' \cap B) = $ value, outside $= 0.1$ | B1 | |
| Need $P(B) = 1 - 0.1 - 0.4 - P(A' \cap B)$; $P(A' \cap B)$ found from $P(B) - 0.3$ | | |
| $P(A' \cap B') = 0.1$, $P(A \cap B') = 0.4$, $P(A \cap B) = 0.3$, $P(A' \cap B) = 0.2$ | A1 ft | All four regions correct |
## Part (iii)
| $P(B) = 0.3 + 0.2 = 0.5$ | M1 | |
| $P(B|A) = \frac{3}{7} \neq P(B) = 0.5$, so not independent | A1 | Must give reason |
## Part (iv)
| $P(B|A) = \frac{3}{7} \approx 0.43 < P(B) = 0.5$ | M1 | |
| Parents attending reduces probability of scoring, so Isobel should ask them not to attend | A1 | |
---2 Isobel plays football for a local team. Sometimes her parents attend matches to watch her play.
\begin{itemize}
\item $A$ is the event that Isobel's parents watch a match.
\item $B$ is the event that Isobel scores in a match.
\end{itemize}
You are given that $\mathrm { P } ( B \mid A ) = \frac { 3 } { 7 }$ and $\mathrm { P } ( A ) = \frac { 7 } { 10 }$.\\
(i) Calculate $\mathrm { P } ( A \cap B )$.
The probability that Isobel does not score and her parents do not attend is 0.1 .\\
(ii) Draw a Venn diagram showing the events $A$ and $B$, and mark in the probability corresponding to each of the regions of your diagram.\\
(iii) Are events $A$ and $B$ independent? Give a reason for your answer.\\
(iv) By comparing $\mathrm { P } ( B \mid A )$ with $\mathrm { P } ( B )$, explain why Isobel should ask her parents not to attend.
\hfill \mbox{\textit{OCR MEI S1 2006 Q2 [8]}}